equivalent formulation of Nakayama’s lemma

The following is equivalent to Nakayama’s lemma.

Let $A$ be a ring, $M$ be a finitely-generated $A$ -module, $N$ a submodule of $M$ , and $\\mathfrak{a}$ an ideal of $A$ contained in its Jacobson radical. Then $M=\\mathfrak{a}M+N\\Rightarrow M=N$ .

Clearly this statement implies Nakayama’s Lemma, by setting $N$ to $0$ . To see that it follows from Nakayama’s Lemma, note first that by the second isomorphism theorem for modules,

\\frac{\\mathfrak{a}M+N}{N}=\\frac{\\mathfrak{a}M}{\\mathfrak{a}M\\cap N}

and the obvious map

\\mathfrak{a}M\\to\\mathfrak{a}\\frac{M}{N}:am\\mapsto a(m+N)

is surjective; the kernel is clearly $\\mathfrak{a}M\\cap N$ . Thus

\\frac{\\mathfrak{a}M+N}{N}\\cong\\mathfrak{a}\\frac{M}{N}

So from $M=\\mathfrak{a}M+N$ we get $M/N=\\mathfrak{a}(M/N)$ . Since $\\mathfrak{a}$ is contained in the Jacobson radical of $M$ , it is contained in the Jacobson radical of $M/N$ , so by Nakayama, $M/N=0$ , i.e. $M=N$ .